CE 595: Finite Elements in Elasticity
HW#1 Solutions
Problem #1
Given: A cantilevered beam has a length L, constant cross-sectional area A, constant moment of inertia
I z , and a variable Young’s modulus E  x   Eo 1  Lx  . It is loaded by a vertically upward line
load w  x   Lo . Assume that there is no body force present and that bending strain energy is the
only significant contributor to internal work.
wx
Required: Using the Rayleigh-Ritz method on the Principle of Virtual Displacements, derive an
approximate expression for the vertical deflection v  x  of the beam’s centerline using the
approximating function v  x   a2 x 2  a3 x3  a4 x 4 , where a2 , a3 , and a4 are unknown constants.
d 2v
, and bending stress is computed by
dx2
write your volume integrals in the form
Hint: Bending strain is computed by  x  x, y    y
 x  x, y   E  x    x, y  .


Be


   dV       dA dx
L
volume
0  area

sure
to
and remember how I z is defined!
Solution: In the absence of body forces, the Principle of Virtual Displacements takes the following form:
 σ nˆ  δudA    σ  δεdV .
surface
volume
In words, the virtual work done by the external forces must equal the virtual strain energy stored in the
beam. Let’s start with the work done by external forces. Since both the applied load w  x  and the
expected displacement v  x  are positive, the real work done is simply the product of w  x  and v  x 
integrated over the beam. Hence, the virtual work is the product of w  x  and  v  x  integrated over the
beam, where the virtual displacement is simply  v  x    a2 x 2   a3 x 3   a4 x 4 , where  a2 ,  a3 , and
 a4 are arbitrary constants. Since w  x  is a line load (and thus is already integrated over the beam’s
thickness), the integral over the surface becomes:

L
surface
surface
wo
L

wo x
*  a2 x 2   a3 x 3   a4 x 4 dx.
0 L
0
 σ   nˆ  δudA 


L
σ   nˆ  δudA   w  x   v  x  dx  
  a x
L
2
3

  a3 x 4   a4 x 5 dx 
0
wo
L
5
4
6
  a2 L  a3 L  a4 L 



.
5
6 
 4
As for the bending strain energy, we follow the information given in the hint:
d 2v
  x    y 2   y 2a2  6a3 x  12a4 x 2 .
dx
2
d v
  x    y
  y 2 a2  6 a3 x  12 a4 x 2 and   x   E  x    x    yEo 1  Lx  2a2  6a3 x  12a4 x 2 .
dx 2
L


   σ  δεdV      yEo 1  Lx  2a2  6a3 x  12a4 x 2 *   y  2 a2  6 a3 x  12 a4 x 2 dA dx
 area
volume
0












y 2 dA  dx.
0
 area

We recognize the integral over the area as the definition of I z , which is a constant in this problem.
Pulling this out and performing the resulting x integral is done in the following Matlab session:
L
 Eo  1 
x
L

  2a2  6a3 x  12a4 x 2  2 a2  6 a3 x  12 a4 x 2   
>> syms x L a2 a3 a4 da2 da3 da4 wo Eo Iz
>> strainenergy = Eo*Iz*expand( int( (1+x/L)*(2*a2 + 6*a3*x +
12*a4*x^2)*(2*da2 + 6*da3*x + 12*da4*x^2),x,0,L))
strainenergy =
Eo*Iz*(264/5*L^5*a4*da4+162/5*L^4*da4*a3+162/5*L^4*a4*da3+14*L^3*da4*a
2+21*L^3*da3*a3+14*L^3*a4*da2+10*L^2*da3*a2+10*L^2*da2*a3+6*a2*da2*L)
This expression minus the external virtual work must equal zero, according to PVD:
>> extwork = wo/L*(da2*L^4/4 + da3*L^5/5 + da4*L^6/6)
extwork =
wo/L*(1/4*da2*L^4+1/5*da3*L^5+1/6*da4*L^6)
>> pvd = strainenergy - extwork;
According to the Rayleigh-Ritz method, we obtain equations to solve for a2 , a3 , and a4 by taking partial
derivatives of the PVD statement with respect to the corresponding virtual coeffcients; i.e.,
   Wi   We  0 
These calculations are shown below:
>> pvd = strainenergy - extwork;
>> eqn1 = diff(pvd,da2)

 0 for k  2,3,4.
 ak
eqn1 =
Eo*Iz*(14*L^3*a4+10*L^2*a3+6*a2*L)-1/4*wo*L^3
>> eqn2 = diff(pvd,da3)
eqn2 =
Eo*Iz*(162/5*L^4*a4+21*L^3*a3+10*L^2*a2)-1/5*wo*L^4
>> eqn3 = diff(pvd,da4)
eqn3 =
Eo*Iz*(264/5*L^5*a4+162/5*L^4*a3+14*L^3*a2)-1/6*wo*L^5
>> consts = solve(eqn1,eqn2,eqn3,a2,a3,a4)
consts =
***
***
a2: [1x1 sym] ***
a3: [1x1 sym] ***
a4: [1x1 sym] ***
consts is an example of a Matlab entity called
a structure. In this case, the solutions for the
unknown constants are the components of this
structure. The way to access the values of these
components is shown in the subsequent commands.
>> consts.a2
ans =
61/378/Eo/Iz*wo*L^2
>> consts.a3
ans =
-263/2268/Eo/Iz*L*wo
>> consts.a4
ans =
143/4536/Eo/Iz*wo
Based on these results, we can write the approximate deflection curve:
61
v  x   378
wo L2
Eo I z
263 o
143
o
x2  2268
x3  4536
x4 . Answer
EI
EI
wL
o z
w
o z
(How good is this solution? Though not required for this problem, you can actually integrate the
governing equation to get the exact solution – as you might guess, it’s a mess, but it’s not too bad in
Matlab. You can then compare the exact and approximate solutions. Below, I show a plot of these two
solutions in nondimensional form:
wo L4
simply relates the size of the load to the stiffness of the beam.) You can see
Eo I z
that the approximate solution is an overestimate of the exact solution, and it’s really only close near x =
0. If you were really interested in getting a good approximation for this problem, this result suggests the
need to look at a higher order approximation – say, adding in a term like a5 x5 to the given approximate
solution.)
(The parameter vo 